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 maxn = 1e6 + 10;
const int mod = 1000000000 + 7;
int dp[2][maxn], nxt[2][maxn];
int a[maxn];
int len[maxn];
int n;
int main() {
  string s;
  cin >> s;
  n = s.size();
  for (int i = 1; i <= n; i++) a[i] = s[i - 1] - '0';
  nxt[0][n + 1] = nxt[1][n + 1] = n + 1;
  for (int i = n; i >= 1; i--) {
    for (int bit = 0; bit <= 1; bit++) {
      nxt[bit][i] = nxt[bit][i + 1];
    }
    nxt[a[i]][i] = i;
  }
  int sum = 0;
  long long total = 1;
  len[0] = 1;
  int cnt = 0;
  for (int i = n; i >= 1; i--) {
    if (a[i] == 0) {
      ++cnt;
    } else
      break;
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == 0) {
      int var = nxt[1][i] - i;
      total += (mod - len[var]);
      len[var] = dp[1][nxt[1][i]];
      if (var <= cnt) len[var] = (len[var] + 1) % mod;
      total += len[var];
      total %= mod;
    }
    dp[0][i] = total;
    if (a[i] == 1) {
      sum += total;
      sum %= mod;
    }
    dp[1][i] = sum;
  }
  if (nxt[1][1] <= n)
    cout << (1LL * dp[1][nxt[1][1]] * nxt[1][1]) % mod << endl;
  else
    cout << n << endl;
  return 0;
}
