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;
const int MAXN = (int)1e6 + 3;
const int infint = (int)1e9 + 3;
const long long inf = (long long)1e18 + 3;
const int MOD = (int)1e9 + 7;
int n, dp[MAXN], part[MAXN], nxt[MAXN];
vector<int> sz;
string s;
int add(int a, int b) {
  int c = a + b;
  if (c < 0) c += MOD;
  if (c >= MOD) c -= MOD;
  return c;
}
int mul(int a, int b) { return (1LL * a * b) % MOD; }
int get(int l, int r) { return add(part[l], -part[r + 1]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  int ind = -1;
  for (int i = 0; i < s.size(); i++)
    if (s[i] == '1') {
      ind = i;
      break;
    }
  if (ind == -1) return cout << n, 0;
  ind += 1;
  int tmp = 0;
  for (int i = ind; i < n; i++)
    if (s[i] == '0')
      tmp++;
    else {
      sz.push_back(tmp);
      tmp = 0;
    }
  sz.push_back(tmp);
  int ptr = sz.size() - 1;
  dp[ptr] = sz.back() + 1;
  part[ptr] = sz.back() + 1;
  for (int i = 0; i <= n; i++) nxt[i] = ptr;
  for (int i = ptr - 1; i >= 0; i--) {
    for (int j = 0; j <= sz[i]; j++) dp[i] = add(dp[i], get(i + 1, nxt[j]));
    part[i] = add(part[i + 1], dp[i]);
    for (int j = 0; j <= sz[i]; j++) nxt[j] = i;
  }
  int ans = 0;
  for (int i = 0; i <= ptr; i++) ans = add(ans, dp[i]);
  cout << mul(ans, ind);
}