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 N = 1e6 + 101;
const int MOD = 1e9 + 7;
int add(int a, int b) { return (a + b) % MOD; }
int n, pref[N], dp[N], kiedy[N];
char s[N];
vector<int> S;
int range(int l, int r) {
  return (pref[r] - (l ? pref[l - 1] : 0) + MOD) % MOD;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int z = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '0') z++;
    if (s[i] == '1') {
      if (z) S.push_back(z);
      S.push_back(0);
      z = 0;
    }
  }
  if (S.empty()) {
    printf("%d\n", n);
    return 0;
  }
  int last = -1;
  for (int i = 0; i <= n; ++i) kiedy[i] = -1;
  int ans = 0;
  for (int i = 0; i <= (int)S.size() - 1; ++i) {
    int u = S[i];
    if (!u) {
      if (last == -1)
        dp[i] = add(1, (i > 0 ? dp[i - 1] : 0));
      else {
        if (!S[i - 1])
          dp[i] = dp[i - 1];
        else
          dp[i] = add(dp[i - 1], dp[i - 2]);
      }
      last = i;
    } else {
      if (!i)
        dp[i] = u;
      else
        for (int j = 1; j <= u; ++j)
          dp[i] = add(dp[i], range(kiedy[j] + 1, i - 1));
      for (int j = 1; j <= u; ++j) kiedy[j] = i;
    }
    if (!u || i + 1 == (int)S.size()) ans = add(ans, dp[i]);
    pref[i] = add(pref[i - 1], (!u ? dp[i] : 0));
  }
  if (z) ans = 1LL * ans * (z + 1) % MOD;
  printf("%d\n", ans);
  return 0;
}
