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 double ONE = 1.0;
const int MAXN = 100005;
const int INF = 0X3F3F3F3F;
int N, Q;
double P[MAXN];
struct Node {
  int l, r;
  pair<double, double> val;
} s[MAXN * 8];
void Push_Up(int rt) {
  pair<double, double> ret1 = s[rt * 2].val, ret2 = s[rt * 2 + 1].val;
  s[rt].val.first =
      ret1.first * ret2.first / (1 - ret1.second * (1 - ret2.first));
  s[rt].val.second = ret2.second + (1 - ret2.second) * ret2.first *
                                       ret1.second /
                                       (1 - ret1.second * (1 - ret2.first));
}
void Build(int rt, int l, int r) {
  s[rt].l = l;
  s[rt].r = r;
  s[rt].val.first = s[rt].val.second = -INF;
  if (l == r) {
    s[rt].val.first = P[l];
    s[rt].val.second = P[l];
    return;
  }
  int mid = (l + r) / 2;
  Build(rt * 2, l, mid);
  Build(rt * 2 + 1, mid + 1, r);
  Push_Up(rt);
}
void Insert(int rt, int p, double val) {
  if (s[rt].l > p || s[rt].r < p) return;
  if (s[rt].l == s[rt].r) {
    s[rt].val.first = s[rt].val.second = val;
    return;
  }
  Insert(rt * 2, p, val);
  Insert(rt * 2 + 1, p, val);
  Push_Up(rt);
}
pair<double, double> Query(int rt, int l, int r) {
  pair<double, double> ret;
  ret.first = ret.second = -INF;
  if (s[rt].l > r || s[rt].r < l) return ret;
  if (s[rt].l >= l && s[rt].r <= r) return s[rt].val;
  pair<double, double> ret1, ret2;
  ret1 = Query(rt * 2, l, r);
  ret2 = Query(rt * 2 + 1, l, r);
  if (ret1.first == -INF && ret2.first == -INF) return ret;
  if (ret1.first == -INF)
    ret = ret2;
  else if (ret2.first == -INF)
    ret = ret1;
  else {
    ret.first = ret1.first * ret2.first / (1 - ret1.second * (1 - ret2.first));
    ret.second = ret2.second + (1 - ret2.second) * ret2.first * ret1.second /
                                   (1 - ret1.second * (1 - ret2.first));
  }
  return ret;
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 1, A, B; i <= N; i++) {
    scanf("%d%d", &A, &B);
    P[i] = ONE * A / B;
  }
  Build(1, 1, N);
  for (int i = 1, k, x, y, z; i <= Q; i++) {
    scanf("%d%d%d", &k, &x, &y);
    if (k == 1) {
      scanf("%d", &z);
      Insert(1, x, ONE * y / z);
    }
    if (k == 2) printf("%.10f\n", Query(1, x, y).first);
  }
  return 0;
}