Koa the Koala has a binary string s of length n. Koa can perform no more than n-1 (possibly zero) operations of the following form:

In one operation Koa selects positions i and i+1 for some i with 1 ≤ i < |s| and sets s_i to max(s_i, s_{i+1}). Then Koa deletes position i+1 from s (after the removal, the remaining parts are concatenated).

Note that after every operation the length of s decreases by 1.

How many different binary strings can Koa obtain by doing no more than n-1 (possibly zero) operations modulo 10^9+7 (1000000007)?

Input

The only line of input contains binary string s (1 ≤ |s| ≤ 10^6). For all i (1 ≤ i ≤ |s|) s_i = 0 or s_i = 1.

Output

On a single line print the answer to the problem modulo 10^9+7 (1000000007).

Examples

Input


000


Output


3


Input


0101


Output


6


Input


0001111


Output


16


Input


00101100011100


Output


477

Note

In the first sample Koa can obtain binary strings: 0, 00 and 000.

In the second sample Koa can obtain binary strings: 1, 01, 11, 011, 101 and 0101. For example:

  * to obtain 01 from 0101 Koa can operate as follows: 0101 → 0(10)1 → 011 → 0(11) → 01. 
  * to obtain 11 from 0101 Koa can operate as follows: 0101 → (01)01 → 101 → 1(01) → 11. 



Parentheses denote the two positions Koa selected in each operation.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MAXN = 1000005;
char s[MAXN];
int n, nxt[MAXN];
long long dp[MAXN], A[MAXN];
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = n; i >= 0; i--) {
    if (i == n)
      nxt[i] = -1;
    else {
      if (s[i] == '1')
        nxt[i] = i;
      else
        nxt[i] = nxt[i + 1];
    }
  }
  int j = -1;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '1') {
      j = i;
      break;
    }
  }
  if (j == -1) {
    printf("%d\n", n);
    return 0;
  }
  int m = j + 1;
  long long r0 = 0;
  for (int i = m; i >= 0; i--) {
    if (i == m)
      dp[i] = 1;
    else {
      int l = nxt[i] - i;
      if (l > 0) {
        r0 = (r0 - A[l] + MOD) % MOD;
        A[l] = dp[nxt[i] + 1];
        r0 = (r0 + A[l]) % MOD;
      }
      dp[i] = (r0 + dp[nxt[i] + 1] + 1) % MOD;
    }
  }
  int k = nxt[0];
  long long res = dp[k + 1] * (n - m + 1) % MOD * (k + 1) % MOD;
  printf("%lld\n", res);
  return 0;
}
