Problem: There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi to lose and move to the casino on the left (i - 1) or also exit the row (if i = 1). 

We say that Memory dominates on the interval i... j if he completes a walk such that,

  * He starts on casino i. 
  * He never looses in casino i. 
  * He finishes his walk by winning in casino j. 



Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process.

Now Memory has some requests, in one of the following forms:

  * 1 i a b: Set <image>. 
  * 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... r after winning at casino r, if she starts in casino l. 



It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. pi ≤ pi + 1 for all i from 1 to n - 1.

Please help Memory by answering all his requests!

Input

The first line of the input contains two integers n and q(1 ≤ n, q ≤ 100 000), — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109) — <image> is the probability pi of winning in casino i.

The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n).

It's guaranteed that there will be at least one query of type 2, i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.

Output

Print a real number for every request of type 2 — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed 10 - 4.

Namely: let's assume that one of your answers is a, and the corresponding answer of the jury is b. The checker program will consider your answer correct if |a - b| ≤ 10 - 4.

Example

Input

3 13
1 3
1 2
2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3
1 2 2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3


Output

0.3333333333
0.2000000000
0.1666666667
0.5000000000
0.4000000000
0.6666666667
0.3333333333
0.2500000000
0.2222222222
0.6666666667
0.5714285714
0.6666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
struct mat {
  long double v[2][2];
  mat() { memset(v, 0, sizeof(v)); }
  mat operator*(const mat &A) const {
    mat ret;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++) ret.v[i][j] += v[i][k] * A.v[k][j];
    return ret;
  }
  void build(long double p) {
    memset(v, 0, sizeof(v));
    v[0][1] = -(1 - p) / p;
    v[1][0] = 1;
    v[1][1] = 1 / p;
  }
} tr[N << 2];
int n, q;
long double p[N];
long double read() {
  int a = gi(), b = gi();
  return 1.0 * a / b;
}
void modify(int p, long double v, int x = 1, int l = 1, int r = n) {
  if (l == r) {
    tr[x].build(v);
    return;
  }
  int mid = (l + r) >> 1;
  p <= mid ? modify(p, v, (x << 1), l, mid)
           : modify(p, v, (x << 1 | 1), mid + 1, r);
  tr[x] = tr[(x << 1)] * tr[(x << 1 | 1)];
}
mat query(int L, int R, int x = 1, int l = 1, int r = n) {
  if (L <= l && r <= R) return tr[x];
  int mid = (l + r) >> 1;
  if (R <= mid) return query(L, R, (x << 1), l, mid);
  if (L > mid) return query(L, R, (x << 1 | 1), mid + 1, r);
  return query(L, R, (x << 1), l, mid) * query(L, R, (x << 1 | 1), mid + 1, r);
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) modify(i, p[i] = read());
  while (q--) {
    int op = gi();
    if (op == 1) {
      int i = gi();
      modify(i, p[i] = read());
    } else {
      int l = gi(), r = gi();
      if (l == r)
        printf("%.10Lf\n", p[l]);
      else {
        auto t = query(l, r);
        mat s;
        s.v[0][1] = 1;
        s = s * t;
        long double ans = 0;
        if (s.v[0][1] < 1e30) ans = 1 / s.v[0][1];
        printf("%.10Lf\n", ans);
      }
    }
  }
  return 0;
}