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 N = 1e6 + 100;
const int mod = 1e9 + 7;
string s;
int n;
vector<pair<int, int>> zeros;
int nxt[N];
int nxt_one[N];
int nxt_zero[N];
int sz[N];
int dp[N];
int solve(int pos, int cnt) {
  if (dp[pos] != -1) return dp[pos];
  int ans = 0;
  if (s[pos] == '1')
    ans = 1;
  else {
    if (s.back() == '0') {
      int x = zeros.back().first;
      if (x >= cnt) {
        if (s[0] == '1' or pos >= zeros[0].first) ans = 1;
      }
    }
  }
  if (s[pos] == '1') {
    if (nxt_one[pos] != -1) ans += solve(nxt_one[pos], 0);
    if (nxt_zero[pos] != -1) ans += solve(nxt_zero[pos], 1);
    if (ans >= mod) ans -= mod;
  } else {
    if (nxt_one[pos] != -1) ans += solve(nxt_one[pos], 0);
    if (pos + 1 < n && s[pos + 1] == '0') ans += solve(nxt_zero[pos], cnt + 1);
    if (pos + 1 < n && s[pos + 1] != '0' && nxt[pos] != -1 &&
        sz[pos] != pos + 1) {
      int l = nxt[pos];
      int idx = l + sz[pos];
      assert(s[idx] == '0');
      ans += solve(idx, cnt + 1);
    }
    ans %= mod;
  }
  return dp[pos] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.size();
  int z = 0;
  for (int i = 0; i < s.size(); i++) z += (s[i] == '0');
  if (z == n) {
    cout << n << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') continue;
    int l = i, r = i;
    for (int j = i; j < n; j++) {
      if (s[j] == '0')
        r = j;
      else
        break;
    }
    zeros.push_back({r - l + 1, l});
    i = r;
  }
  if (zeros.size() == 0) {
    cout << n << endl;
    return 0;
  }
  stack<pair<int, int>> stk;
  for (int i = zeros.size() - 1; i >= 0; i--) {
    while (stk.size() > 0 && stk.top().first <= zeros[i].first) stk.pop();
    int r = zeros[i].second + zeros[i].first - 1;
    nxt[r] = -1;
    sz[r] = zeros[i].first;
    if (stk.size() > 0) nxt[r] = stk.top().second;
    stk.push(zeros[i]);
  }
  int last = -1;
  int last0 = -1;
  for (int i = n - 1; i >= 0; i--) {
    nxt_one[i] = last;
    nxt_zero[i] = last0;
    if (s[i] == '1')
      last = i;
    else
      last0 = i;
  }
  memset(dp, -1, sizeof dp);
  int ans = 0;
  if (s[0] == '1') {
    ans = solve(0, 0);
  } else {
    ans = solve(0, 1);
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        ans += solve(i, 0);
        break;
      }
    }
    ans %= mod;
  }
  cout << ans << endl;
}