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 MOD = 1000000007;
const int INF = 2000000000;
const long long BIG = 1446803456761533460LL;
const int maxn = 1000010;
string second;
int n, dp[maxn], ps, stk[maxn], stkn;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> second;
  n = second.size();
  int i = 0;
  while (i < n && second[i] == '0') ++i;
  if (i == n) return cout << n << '\n', 0;
  ps = i + 1;
  while (i < n) {
    int j = i + 1;
    while (j < n && second[j] == '0') ++j;
    int z = j - i;
    int v = ps;
    while (stkn && stk[stkn - 1] <= z) {
      ps = ((ps) +
            (MOD - 1LL * dp[stkn - 1] * (stk[stkn - 1] - stk[stkn]) % MOD)) %
           MOD;
      --stkn;
    }
    stk[stkn + 1] = 0;
    if (stkn)
      ps = ((ps) +
            (MOD - 1LL * dp[stkn - 1] * (stk[stkn - 1] - stk[stkn]) % MOD)) %
           MOD;
    stk[stkn] = z;
    if (stkn)
      ps = ((ps) + (1LL * dp[stkn - 1] * (stk[stkn - 1] - stk[stkn] % MOD))) %
           MOD;
    dp[stkn] = v;
    ps = ((ps) + (1LL * v * z % MOD)) % MOD;
    ++stkn;
    i = j;
  }
  int ans = 1LL * dp[stkn - 1] * stk[stkn - 1] % MOD;
  cout << ans << '\n';
}
