#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e5 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long n, a[105], b[105], M[N], ans;
string s;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> s;
  for (long long i = 1; i <= n; i++) cin >> a[i] >> b[i];
  for (long long i = 1; i <= n; i++) {
    long long x = s[i - 1] - '0';
    M[0] += x;
    for (long long j = 0; j <= (1e5 - b[i]) / a[i]; j++) {
      if (!x)
        M[j * a[i] + b[i]]++;
      else
        M[j * a[i] + b[i]]--;
      x ^= 1;
    }
  }
  ans = M[0];
  for (long long i = 1; i <= 1e5; i++)
    M[i] += M[i - 1], ans = (M[i] > ans ? M[i] : ans);
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e5 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long n, a[105], b[105], M[N], ans;
string s;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> s;
  for (long long i = 1; i <= n; i++) cin >> a[i] >> b[i];
  for (long long i = 1; i <= n; i++) {
    long long x = s[i - 1] - '0';
    M[0] += x;
    for (long long j = 0; j <= (1e5 - b[i]) / a[i]; j++) {
      if (!x)
        M[j * a[i] + b[i]]++;
      else
        M[j * a[i] + b[i]]--;
      x ^= 1;
    }
  }
  ans = M[0];
  for (long long i = 1; i <= 1e5; i++)
    M[i] += M[i - 1], ans = (M[i] > ans ? M[i] : ans);
  cout << ans;
  return 0;
}

```