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;
int n, q;
struct mtrx {
  double A[2][2];
  mtrx() { memset(A, 0, sizeof A); }
  mtrx(double x) {
    A[0][0] = 1. / x, A[0][1] = (x - 1.) / x, A[1][0] = 1., A[1][1] = 0;
  }
  mtrx operator*(const mtrx B) {
    mtrx ret;
    for (int i = (0), iend = (2 - 1); i <= iend; ++i)
      for (int j = (0), jend = (2 - 1); j <= jend; ++j)
        for (int k = (0), kend = (2 - 1); k <= kend; ++k)
          ret.A[i][j] += A[i][k] * B.A[k][j];
    return ret;
  }
  void ot() {
    for (int i = 0; i < 2; ++i) {
      for (int j = (0), jend = (1); j <= jend; ++j) printf("%lf ", A[i][j]);
      puts("");
    }
  }
} T[200026 << 2];
double p[200026];
void pu(int rt) { T[rt] = T[rt << 1 | 1] * T[rt << 1]; }
void build(int rt, int l, int r) {
  if (l == r) {
    T[rt] = mtrx(p[l]);
    return;
  }
  int m = l + r >> 1;
  build(rt << 1, l, m), build(rt << 1 | 1, m + 1, r);
  pu(rt);
}
mtrx que(int rt, int l, int r, int L, int R) {
  if (L <= l && R >= r) {
    return T[rt];
  }
  int m = l + r >> 1;
  mtrx re;
  re.A[0][0] = re.A[1][1] = 1;
  if (R > m) re = re * que(rt << 1 | 1, m + 1, r, L, R);
  if (L <= m) re = re * que(rt << 1, l, m, L, R);
  return re;
}
void mdfy(int rt, int l, int r, int p, double c) {
  if (l == r) {
    T[rt] = mtrx(c);
    return;
  }
  int m = l + r >> 1;
  if (p <= m)
    mdfy(rt << 1, l, m, p, c);
  else
    mdfy(rt << 1 | 1, m + 1, r, p, c);
  pu(rt);
}
void solve() {
  cin >> n >> q;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    p[i] = 1. * a / b;
  }
  build(1, 1, n);
  for (int i = (1), iend = (q); i <= iend; ++i) {
    int op, l, r, a, b;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &l, &a, &b);
      p[l] = 1. * a / b;
      mdfy(1, 1, n, l, p[l]);
    } else {
      scanf("%d%d", &l, &r);
      if (l == r) {
        printf("%.8lf\n", p[l]);
        continue;
      }
      mtrx t;
      t.A[0][0] = 1. / p[l], t.A[1][0] = 1.;
      mtrx re = que(1, 1, n, l + 1, r);
      re = re * t;
      double as = 1. / re.A[0][0];
      if (isnan(as))
        puts("0.00000000");
      else
        printf("%.8lf\n", 1. / re.A[0][0]);
    }
  }
}
signed main() { solve(); }