Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples

Input

3 4 1 1


Output

7


Input

3 4 2 1


Output

25


Input

3 4 3 1


Output

79
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 5e5, mo = 1e9 + 7;
typedef long long matrix[5][5];
typedef long long vec[5];
matrix a;
vec f;
long long A, B, n, x, i, j, k;
void mul(matrix a, matrix b, matrix res) {
  matrix c;
  memset(c, 0, sizeof(c));
  long long i, j, k;
  for (i = 1; i <= 2; i++)
    for (j = 1; j <= 2; j++)
      for (k = 1; k <= 2; k++) c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mo;
  memcpy(res, c, sizeof(c));
}
void ksm(matrix a, long long b, matrix res) {
  matrix t = {0};
  long long i;
  for (i = 1; i <= 2; i++) t[i][i] = 1;
  while (b) {
    if (b & 1) mul(t, a, t);
    mul(a, a, a);
    b >>= 1;
  }
  memcpy(res, t, sizeof(t));
}
void mul(vec f, matrix a, vec res) {
  vec c;
  memset(c, 0, sizeof(c));
  long long i, j;
  for (i = 1; i <= 2; i++)
    for (j = 1; j <= 2; j++) c[i] = (c[i] + f[j] * a[j][i]) % mo;
  memcpy(res, c, sizeof(c));
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &A, &B, &n, &x);
  f[1] = x;
  f[2] = 1;
  a[1][1] = A;
  a[2][1] = B;
  a[2][2] = 1;
  ksm(a, n, a);
  mul(f, a, f);
  printf("%I64d", f[1]);
  return 0;
}
