Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, a[1000010], fir, lst, f[1000010][2], c[1000010], dp[1000010][2];
string h;
long long s[1000010], top;
void add(long long &x, long long y) {
  x = x + y;
  x %= 1000000007;
}
signed main() {
  cin >> h;
  n = h.size();
  for (long long i = 1; i <= n; i++) {
    a[i] = h[i - 1] - '0';
    if (a[i] == 1) {
      if (!fir) fir = i;
      lst = max(lst, i);
    }
  }
  if (!fir) {
    cout << n << endl;
    return 0;
  }
  for (long long i = 0; i <= n + 5; i++)
    for (long long j = 0; j <= 1; j++) f[i][j] = n + 1;
  for (long long i = n; i >= 1; i--) {
    if (a[i + 1] == 1)
      f[i][1] = i + 1;
    else
      f[i][1] = f[i + 1][1];
  }
  long long now = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 1)
      c[i - 1] = now, now = 0;
    else {
      now++;
      for (long long j = i - 1; j >= 1; j--) {
        f[j][0] = i;
        if (!a[j]) break;
      }
    }
  }
  if (a[n] == 0) c[n] = now;
  c[n + 1] = 1 << 30, s[++top] = n + 1;
  for (long long i = n; i >= 1; i--) {
    if (!c[i]) continue;
    while (c[i] >= c[s[top]]) top--;
    if (top == 1)
      f[i][0] = n + 1;
    else
      f[i][0] = s[top] - c[s[top]] + c[i] + 1;
    s[++top] = i;
  }
  dp[fir][1] = 1;
  long long ans = 0;
  for (long long i = fir; i <= lst; i++) {
    for (long long j = 0; j <= 1; j++)
      for (long long k = 0; k <= 1; k++) add(dp[f[i][k]][k], dp[i][j]);
    add(ans, dp[i][1]);
  }
  cout << ans * fir % 1000000007 * (n - lst + 1) % 1000000007 << endl;
}