#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long p2(long long n) { return (1ll << n); }
long long getbit(long long j, long long i) { return ((i >> (j - 1ll)) & 1ll); }
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1000000007;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
const double eps = 1e-7;
const double pi = acos(-1.0);
inline void inc(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline void dec(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
inline void chmax(long long &x, long long y) {
  if (y > x) x = y;
}
inline void mulm(long long &x, long long y) {
  x *= y;
  if (x >= mod) x -= mod;
}
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
vector<long long> tree[100005];
long long a[100005];
void build(long long node, long long l, long long r) {
  if (l == r) {
    tree[node].push_back(a[l]);
    return;
  }
  long long m = (l + r) / 2ll;
  build(node * 2, l, m);
  build(node * 2 + 1, m + 1, r);
  tree[node].resize(tree[node * 2].size() + tree[2 * node + 1].size());
  merge(tree[node * 2].begin(), tree[node * 2].end(),
        tree[node * 2 + 1].begin(), tree[node * 2 + 1].end(),
        tree[node].begin());
}
long long query(long long node, long long l, long long r, long long st,
                long long en, long long v) {
  if (l > en || r < st) return 0;
  if (l >= st && r <= en) {
    return tree[node].end() -
           lower_bound((tree[node]).begin(), (tree[node]).end(), v);
  }
  long long m = (l + r) / 2ll;
  return query(node * 2, l, m, st, en, v) +
         query(node * 2 + 1, m + 1, r, st, en, v);
}
int main() {
  long long t, n, k, m, j;
  cin >> n;
  long long lasz = 0, c = 0;
  for (long long i = 1; i <= n; i++) {
    char ch;
    cin >> ch;
    if (ch == '1')
      c++;
    else {
      cout << c;
      c = 0;
    }
  }
  cout << c;
}

### Prompt

Generate a cpp solution to the following problem:

Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:

  * Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). 
  * Digits are written one by one in order corresponding to number and separated by single '0' character. 



Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.

Input

The first line contains one integer number n (1 ≤ n ≤ 89) — length of the string s.

The second line contains string s — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.

Output

Print the decoded number.

Examples

Input

3
111


Output

3


Input

9
110011101


Output

2031

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long p2(long long n) { return (1ll << n); }
long long getbit(long long j, long long i) { return ((i >> (j - 1ll)) & 1ll); }
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1000000007;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
const double eps = 1e-7;
const double pi = acos(-1.0);
inline void inc(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline void dec(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
inline void chmax(long long &x, long long y) {
  if (y > x) x = y;
}
inline void mulm(long long &x, long long y) {
  x *= y;
  if (x >= mod) x -= mod;
}
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
vector<long long> tree[100005];
long long a[100005];
void build(long long node, long long l, long long r) {
  if (l == r) {
    tree[node].push_back(a[l]);
    return;
  }
  long long m = (l + r) / 2ll;
  build(node * 2, l, m);
  build(node * 2 + 1, m + 1, r);
  tree[node].resize(tree[node * 2].size() + tree[2 * node + 1].size());
  merge(tree[node * 2].begin(), tree[node * 2].end(),
        tree[node * 2 + 1].begin(), tree[node * 2 + 1].end(),
        tree[node].begin());
}
long long query(long long node, long long l, long long r, long long st,
                long long en, long long v) {
  if (l > en || r < st) return 0;
  if (l >= st && r <= en) {
    return tree[node].end() -
           lower_bound((tree[node]).begin(), (tree[node]).end(), v);
  }
  long long m = (l + r) / 2ll;
  return query(node * 2, l, m, st, en, v) +
         query(node * 2 + 1, m + 1, r, st, en, v);
}
int main() {
  long long t, n, k, m, j;
  cin >> n;
  long long lasz = 0, c = 0;
  for (long long i = 1; i <= n; i++) {
    char ch;
    cin >> ch;
    if (ch == '1')
      c++;
    else {
      cout << c;
      c = 0;
    }
  }
  cout << c;
}

```