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;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long mod5 = 1000000009;
long long inf = 1LL << 61;
int iinf = 1 << 30;
double pi = 3.14159265358979323846;
double pi2 = pi / 2.0;
double eps = 1e-8;
int dh[4] = {1, 0, -1, 0};
int dw[4] = {0, 1, 0, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long gcd(long long a, long long b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return a * b / c;
}
long long Pow(long long n, long long k) {
  if (k < 0) return 0;
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  now %= md;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
long long popcount(long long n) {
  long long ret = 0;
  long long u = n;
  while (u > 0) {
    ret += u % 2;
    u /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = (int)s.size();
  long long cnt = 0;
  vector<long long> v;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0')
      cnt++;
    else {
      v.push_back(cnt + 1);
      cnt = 0;
    }
  }
  if (v.size() == 0) {
    cout << n << endl;
    return 0;
  }
  long long ans = (cnt + 1) * v[0] % mod;
  v.erase(v.begin());
  long long m = v.size();
  long long dp[m + 1];
  dp[0] = 1;
  deque<pair<long long, long long> > dq;
  dq.push_back({1, 0});
  long long anss = 1;
  for (int i = 0; i < m; i++) {
    long long wa = 0;
    dp[i + 1] = 0;
    while (dq.size() > 0 && dq.back().second <= v[i]) {
      (dp[i + 1] += (v[i] - dq.back().second) * dq.back().first) %= mod;
      (wa += dq.back().first) %= mod;
      dq.pop_back();
    }
    dq.push_back({wa, v[i]});
    dq.push_back({dp[i + 1], 0});
    (anss += dp[i + 1]) %= mod;
  }
  (ans *= anss) %= mod;
  cout << ans << endl;
}
