#include <bits/stdc++.h>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool prime[10000001];
long long P = 1000000007 - 2;
long long factorial[10000001] = {0};
long long powerFunction(long long x, long long y) {
  long long res = 1;
  long long p = 1000000007;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void sieveFunction(long long maxLimit) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long i = 2; maxLimit >= i; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; maxLimit >= j; j += i) prime[j] = false;
    }
  }
}
void factorialFunction(long long maxLimit) {
  factorial[0] = 1;
  for (long long i = 1; i <= maxLimit; i++)
    factorial[i] = (factorial[i - 1] * i) % 1000000007;
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  long long n;
  cin >> n;
  string str;
  cin >> str;
  long long t[20001];
  long long a[n];
  long long b[n];
  for (long long i = 0; n > i; i++) {
    cin >> a[i] >> b[i];
  }
  memset(t, 0, sizeof(t));
  for (long long i = 0; n > i; i++) {
    if (str[i] == '1') {
      for (long long j = 0; b[i] > j; j++) {
        t[j]++;
      }
      for (long long j = b[i] + a[i]; 10001 > j; j += 2 * a[i]) {
        for (long long k = 0; a[i] > k; k++) {
          t[j + k]++;
        }
      }
    } else {
      for (long long j = b[i]; 10001 > j; j += 2 * a[i]) {
        for (long long k = 0; a[i] > k; k++) {
          t[j + k]++;
        }
      }
    }
  }
  long long ans = INT_MIN;
  for (long long i = 0; 1001 > i; i++) {
    ans = max(ans, t[i]);
  }
  cout << ans << "\n";
}

### Prompt

Develop a solution in CPP to the problem described below:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool prime[10000001];
long long P = 1000000007 - 2;
long long factorial[10000001] = {0};
long long powerFunction(long long x, long long y) {
  long long res = 1;
  long long p = 1000000007;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void sieveFunction(long long maxLimit) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long i = 2; maxLimit >= i; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; maxLimit >= j; j += i) prime[j] = false;
    }
  }
}
void factorialFunction(long long maxLimit) {
  factorial[0] = 1;
  for (long long i = 1; i <= maxLimit; i++)
    factorial[i] = (factorial[i - 1] * i) % 1000000007;
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  long long n;
  cin >> n;
  string str;
  cin >> str;
  long long t[20001];
  long long a[n];
  long long b[n];
  for (long long i = 0; n > i; i++) {
    cin >> a[i] >> b[i];
  }
  memset(t, 0, sizeof(t));
  for (long long i = 0; n > i; i++) {
    if (str[i] == '1') {
      for (long long j = 0; b[i] > j; j++) {
        t[j]++;
      }
      for (long long j = b[i] + a[i]; 10001 > j; j += 2 * a[i]) {
        for (long long k = 0; a[i] > k; k++) {
          t[j + k]++;
        }
      }
    } else {
      for (long long j = b[i]; 10001 > j; j += 2 * a[i]) {
        for (long long k = 0; a[i] > k; k++) {
          t[j + k]++;
        }
      }
    }
  }
  long long ans = INT_MIN;
  for (long long i = 0; 1001 > i; i++) {
    ans = max(ans, t[i]);
  }
  cout << ans << "\n";
}

```