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 long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  int n = s.length();
  int prezeroes = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') break;
    prezeroes++;
  }
  if (prezeroes == n) {
    cout << n << '\n';
    exit(0);
  }
  int runs[n];
  for (int i = 0; i < n; i++) {
    if (i == 0)
      runs[i] = (s[i] == '0');
    else if (s[i] == '0')
      runs[i] = 1 + runs[i - 1];
    else
      runs[i] = 0;
  }
  int dp[n];
  int seek[n + 1];
  memset(seek, -1, sizeof seek);
  memset(dp, 0, sizeof dp);
  for (int i = n - 1; i >= 0; i--) {
    int currrun = runs[i];
    int a = seek[0];
    int b = seek[currrun + 1];
    if (a != -1) {
      dp[i] = (dp[i] + dp[a]) % MOD;
    }
    if (b != -1) {
      dp[i] = (dp[i] + dp[b]) % MOD;
    }
    if (s[i] == '1') {
      dp[i] = (dp[i] + 1) % MOD;
    } else if (runs[i] <= runs[n - 1]) {
      dp[i] = (dp[i] + 1) % MOD;
    }
    seek[currrun] = i;
  }
  int ans = (1ll * (prezeroes + 1) * dp[prezeroes]) % MOD;
  cout << ans << '\n';
}
