#include "bits/stdc++.h" #define REP(i, n, N) for(ll i=(n); i<(N); i++) #define RREP(i, n, N) for(ll i=(N-1); i>=n; i--) #define CK(n, a, b) ((a)<=(n)&&(n)<(b)) #define ALL(v) (v).begin(),(v).end() #define MCP(a,b) memcpy(b,a,sizeof(b)) #define p(s) cout<<(s)<> typedef long long ll; using namespace std; const ll mod = 1e9 + 7; const ll inf = 1e18; typedef vector vec; typedef vector mat; ll N,C; //A*B mat mul(mat &A, mat &B){ mat C(A.size(), vec(B[0].size(),0)); REP(i, 0, A.size()) REP(k, 0, B.size()) REP(j, 0, B[0].size()) C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod; return C; } //A^n mat mat_pow(mat A, long long n){ mat B(A.size(), vec(A.size(), 0)); REP(i, 0, A.size()) B[i][i] = 1; while(n > 0){ if(n & 1) B = mul(B, A); A = mul(A, A); n >>= 1; } return B; } long long fast_power(long long a, long long b) { long long ret = 1; // aが負の場合 if (a < 0) { a = -a; if (b % 2 == 1) ret = -1; } // 累乗計算 while (b>0) { if (b & 1) { ret *= a; } a *= a; a %= mod; ret %= mod; b >>= 1; } return ret; } string D; int main(){ cin>>N; ll ans=1; mat A(2, vec(2,0)); A[0][0] = A[1][0] = A[0][1] = 1; A[1][1] = 0; REP(i,0,N){ cin>>C>>D; mat AA = mat_pow(A, C+1); mat B(2, vec(1,0)); B[0][0] = 1, B[1][0] = 0; mat F = mul(AA, B); ll d=0; REP(i,0,D.size()){ d *= 10; d += D[i] - '0'; d %= mod; } ans *= fast_power(F[0][0] , d) % mod; ans %= mod; } p(ans); return 0; }