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 MAX = 1e6 + 5, mod = 1e9 + 7, inf = 0x3f3f3f3f;
using ll = long long;
int now;
char s[MAX];
int a[MAX], D[MAX];
ll dp[MAX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s + 1;
  int n = strlen(s + 1);
  int z = 0, dem = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '0') {
      ++dem;
    } else {
      a[++z] = dem;
      dem = 0;
    }
  }
  if (dem == n) {
    cout << n;
    return 0;
  }
  a[++z] = dem;
  dp[z] = 1;
  int top = 0;
  ll ans = a[z] + 1;
  a[0] = -1;
  for (int i = z - 1; i > 0; i--) {
    dp[i] = ans;
    D[top + 1] = 0;
    while (top > 0 && a[D[top]] <= a[i]) {
      ans -= dp[D[top]] * (a[D[top]] - a[D[top + 1]]);
      ans %= mod;
      top--;
    }
    ans = (ans + dp[D[top]] * (a[D[top + 1]] - a[i])) % mod;
    if (ans < 0) ans += mod;
    D[++top] = i;
    ans = (ans + dp[D[top]] * (a[D[top]] + 1)) % mod;
  }
  cout << (dp[1] * (a[1] + 1)) % mod;
}
