#include <bits/stdc++.h>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a + b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long mul(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a * b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
  long long res = 1LL;
  a = a % m;
  while (b) {
    if (b & 1) res = mul(res, a, m);
    b >>= 1;
    a = mul(a, a, m);
  }
  return res;
}
long long fastexp(long long a, long long b) {
  long long res = 1LL;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtendido(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverso(int a, int m) {
  int x, y;
  int g = gcdExtendido(a, m, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % m + m) % m;
}
const int N = 100 + 5;
const int MAX = 1e5;
int n;
string s;
int a[N];
int b[N];
int ac[MAX];
void propagate(int pos) {
  int val = s[pos - 1] - '0';
  for (int i = 0; i < b[pos]; i++) {
    ac[i] += val;
  }
  for (int i = b[pos]; i < MAX; i++) {
    if ((i - b[pos]) % a[pos] == 0) val ^= 1;
    ac[i] += val;
  }
}
int main() {
  scanf("%d", &(n));
  cin >> s;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &(a[i]), &(b[i]));
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) propagate(i);
  ans = *max_element(ac, ac + MAX);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a + b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long mul(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a * b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
  long long res = 1LL;
  a = a % m;
  while (b) {
    if (b & 1) res = mul(res, a, m);
    b >>= 1;
    a = mul(a, a, m);
  }
  return res;
}
long long fastexp(long long a, long long b) {
  long long res = 1LL;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtendido(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverso(int a, int m) {
  int x, y;
  int g = gcdExtendido(a, m, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % m + m) % m;
}
const int N = 100 + 5;
const int MAX = 1e5;
int n;
string s;
int a[N];
int b[N];
int ac[MAX];
void propagate(int pos) {
  int val = s[pos - 1] - '0';
  for (int i = 0; i < b[pos]; i++) {
    ac[i] += val;
  }
  for (int i = b[pos]; i < MAX; i++) {
    if ((i - b[pos]) % a[pos] == 0) val ^= 1;
    ac[i] += val;
  }
}
int main() {
  scanf("%d", &(n));
  cin >> s;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &(a[i]), &(b[i]));
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) propagate(i);
  ans = *max_element(ac, ac + MAX);
  printf("%d\n", ans);
  return 0;
}

```