//113 - Power of Cryptography (by Snail)
#include <iostream>
#include <cmath>
using namespace std;

int main() {
    double n, p;
    while (cin >> n >> p)
        cout << int(pow(p, 1/n)+.5) << endl;
}

//155 - All Squares (by Snail)
#include <iostream>
#include <iomanip>
#include <cstdlib>      // for abs(int)
using namespace std;

int main () {
    int k, x, y, x1, y1, ns;
    while (cin >> k >> x >> y, k) {
        ns = 0;                                 // no of s(quare) -- 正方形數
        x1 = y1 = 1024;                         // 中心點
        while (k) {
            if (abs(x-x1)<=k && abs(y-y1)<=k)   // 在正方形內
                ns++;
            x1 +=  k * ((x >= x1) - (x < x1));   // (x1, y1) 移到最接近的角
            y1 +=  k * ((y >= y1) - (y < y1));
            k /= 2;                             // 縮小正方形
        }
        cout << setw(3) << ns << endl;
    }
}

//10026 - Shoemaker's Problem (by Snail)
#include <iostream>
#include <algorithm>
using namespace std;

struct job {                                    // i(ndex) -- 工作編號
    int i, t, s;                                // t(ime)-- 工作天數
} j[1001];                                      // s (fine) -- 罰金

bool gt (job a, job b) { return a.s * b.t > b.s * a.t; }
                                // 罰得較重 (s/ t 較大) 的先做，交叉相乘
int main () {
    int tc, n, i;
    cin >> tc;
    while (tc--) {
        cin >> n;
        for (i=1; i<=n; i++) {
            j[i].i = i;
            cin >> j[i].t >> j[i].s;
        }
        stable_sort (j+1, j+1+n, gt);           // 穩定排序
        for (i=1; i<n; i++)
            cout << j[i].i << ' ';
        cout << j[n].i << endl;                 // 最後一筆後面沒有空白
        if (tc) cout << endl;                   // 測試資料間空一行
    }
}

//10048 - Audiophobia (by Snail)
#include <iostream>
#include <climits>
using namespace std;


int main() {
    int d[101][101], c, s, q, i, k, c1, c2, tc=0;
    while (cin >> c >> s >> q, c) {
        if (tc++) cout << endl;
        cout << "Case #" << tc << endl;
        for (c1=1; c1<=c; c1++)                 // 歸零
            for (c2=1; c2<=c; c2++)
                d[c1][c2] = INT_MAX;
        for (i=1; i<=s; i++) {
            cin >> c1 >> c2;                    // 輸入街道起訖點
            cin >> d[c1][c2];                   // 輸入街道噪音值
            d[c2][c1] = d[c1][c2];              // 無向圖
        }
        for (k=1; k<=c; k++)                    // Floyd-Warshall
            for (c1=1; c1<=c; c1++)
                for (c2=1; c2<=c; c2++)
                    d[c1][c2] = min(d[c1][c2], max(d[c1][k], d[k][c2]));
        for (i=1; i<=q; i++) {
            cin >> c1 >> c2;
            if (d[c1][c2] == INT_MAX)
                cout << "no path\n";
            else
                cout << d[c1][c2] << endl;
        }
    }
}

//166 - Making Change (by Snail)
#include <iostream>
#include <iomanip>
using namespace std;

int main () {
    int i, j, d, c, a, p[6], nc[201]={};
    int f[6]={5, 10, 20, 50, 100, 200};         // f(ace value) -- 面額
    char ch;
    for (i=1; i<=200; i++)                      // no of coins of change
        nc[i] = 9999;               // nc[i] -- 店家找 i 分時最少要幾個硬幣
    for (i=0; i<6; i++)
        for (j=f[i]; j<=200; j++)   // 硬幣面額最大 $2, 找零不可能大於 $2
            nc[j] = min (nc[j], nc[j-f[i]]+1);
    while (cin >> p[0] >> p[1] >> p[2] >> p[3] >> p[4] >> p[5], p[0]+p[1]+p[2]+p[3]+p[4]+p[5]) {
        int mnc=9999, np[701]={};               // no of coins of payment
        cin >> d >> ch >> c;                    // d(ollar), c(ent)
        a = d*100 + c;                          // a(mount) -- 金額
        for (i=a+200; i; i--)                   // no of coins of payment
            np[i] = 9999;           // np[i] -- 顧客付 i 分時最少要幾個硬幣
        for (i=0; i<6; i++)                     // 面額
            while (p[i]--)                      // 硬幣數量
                for (j=a+200; j>=f[i]; j--)     // 每個硬幣用一次, 倒著 DP
                    np[j] = min (np[j], np[j-f[i]]+1);
        for (i=0; i<200; i++)                   // 溢付金額不超過 $2
            mnc = min (mnc, nc[i]+np[a+i]);
        cout << setw(3) << mnc << endl;
    }
}

//111 - History Grading (by Snail)
#include <iostream>
using namespace std;

int main () {
    int n, i, j, x, c[21], r[21], a[21][21]={};
    cin >> n;
    for (i=1; i<=n; i++)
        cin >> x, c[x] = i;                     // 轉成事件順序
    while (cin >> x) {
        r[x] = 1;
        for (i=2; i<=n; i++)
            cin >> x, r[x] = i;                 // 轉成事件順序
        for (i=1; i<=n; i++)                    // LCS
            for (j=1; j<=n; j++)
                if (c[i]==r[j])
                    a[i][j] = a[i-1][j-1] + 1;
                else
                    a[i][j] = max (a[i][j-1], a[i-1][j]);
        cout << a[n][n] << endl;
    }
}