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 unsigned long long p = 1e9 + 7;
char s[1010000];
int n1[1010000];
int n0[1000100];
int dis[1010000];
int pos[1010000];
unsigned long long dp[1010000];
int main() {
  scanf("%s", s + 1);
  int len = strlen(s + 1), ok = 0;
  n1[len + 1] = n0[len + 1] = pos[len + 1] = dis[len + 1] = len + 1;
  for (int i = 1; i <= len && !ok; i++) {
    ok = (s[i] == '1') * i;
  }
  if (!ok) {
    printf("%d\n", len);
    return 0;
  }
  for (int i = 1; i <= len; i++) {
    pos[i] = len + 1;
    dis[i] = (s[i] == '1' ? 0 : dis[i - 1] + 1);
  }
  for (int i = len; i >= 1; i--) {
    n1[i] = (s[i + 1] == '1' ? i + 1 : n1[i + 1]);
    n0[i] = pos[dis[i] + 1];
    pos[dis[i]] = i;
  }
  unsigned long long ans = 0;
  dp[ok] = 1;
  for (int i = 1; i <= len; i++) {
    dp[n1[i]] = (dp[n1[i]] + dp[i]) % p;
    dp[n0[i]] = (dp[n0[i]] + dp[i]) % p;
    if (dis[i] <= dis[len]) ans = (ans + dp[i]) % p;
  }
  printf("%lld\n", ans * ok % p);
}
