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 = 1000005;
const int MOD = 1000000007;
int n, curr, ima, br;
string s;
int dp[MAXN];
vector<pair<int, int> > v;
vector<int> r;
int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
int mul(int x, int y) { return (long long)x * y % MOD; }
void makni() {
  int len = v.back().first, pos = v.back().second;
  curr = sub(curr, dp[pos]);
  v.pop_back();
  if (!v.empty()) {
    curr = sub(curr, dp[v.back().second - len - 1]);
    curr = add(curr, dp[v.back().second]);
  }
}
void ubaci(int len, int pos) {
  while (!v.empty() && v.back().first <= len) makni();
  if (!v.empty()) {
    curr = sub(curr, dp[v.back().second]);
    curr = add(curr, dp[v.back().second - len - 1]);
  }
  v.push_back({len, pos});
  curr = add(curr, dp[pos]);
}
void calc(int lo, int hi) {
  dp[lo] = add(br + 1, curr);
  int novi = curr;
  if (!v.empty())
    novi = add(sub(novi, dp[v.back().second]), dp[v.back().second - 1]);
  int a = 0, b = 0;
  for (int i = lo + 1; i <= hi; i++) {
    if (s[i] == '1') {
      a++;
      dp[i] = add(dp[i - 1], add(novi, br + 1));
    } else {
      b++;
      dp[i] = mul(dp[lo + a], (b + 1));
    }
  }
  ubaci(b, hi);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    if (s[i] == '0' && ima == 0) br++;
    if (s[i] == '1') ima = 1;
    if (s[i] == '1' && (i == 0 || s[i - 1] == '0')) r.push_back(i);
  }
  if (ima == 0) {
    cout << n;
    return 0;
  }
  r.push_back(n);
  for (int i = 1; i < r.size(); i++) {
    calc(r[i - 1], r[i] - 1);
  }
  cout << dp[n - 1];
  return 0;
}
