Nagini, being a horcrux You-know-who created with the murder of Bertha Jorkins, has accumulated its army of snakes and is launching an attack on Hogwarts school. 

Hogwarts' entrance can be imagined as a straight line (x-axis) from 1 to 105. Nagini is launching various snakes at the Hogwarts entrance. Each snake lands parallel to the entrance, covering a segment at a distance k from x = l to x = r. Formally, each snake can be imagined as being a line segment between points (l, k) and (r, k). Note that k can be both positive and negative, but not 0.

Let, at some x-coordinate x = i, there be snakes at point (i, y1) and point (i, y2), such that y1 > 0 and y2 < 0. Then, if for any point (i, y3) containing a snake such that y3 > 0, y1 ≤ y3 holds and for any point (i, y4) containing a snake such that y4 < 0, |y2| ≤ |y4| holds, then the danger value at coordinate x = i is y1 + |y2|. If no such y1 and y2 exist, danger value is 0. 

Harry wants to calculate the danger value of various segments of the Hogwarts entrance. Danger value for a segment [l, r) of the entrance can be calculated by taking the sum of danger values for each integer x-coordinate present in the segment.

Formally, you have to implement two types of queries:

  * 1 l r k: a snake is added parallel to entrance from x = l to x = r at y-coordinate y = k (l inclusive, r exclusive). 
  * 2 l r: you have to calculate the danger value of segment l to r (l inclusive, r exclusive). 

Input

First line of input contains a single integer q (1 ≤ q ≤ 5·104) denoting the number of queries.

Next q lines each describe a query. Each query description first contains the query type typei (1 ≤ typei ≤ 2). This is followed by further description of the query. In case of the type being 1, it is followed by integers li, ri and ki (<image>,  - 109 ≤ ki ≤ 109, k ≠ 0). Otherwise, it just contains two integers, li and ri (1 ≤ li < ri ≤ 105).

Output

Output the answer for each query of type 2 in a separate line.

Examples

Input

3
1 1 10 10
1 2 4 -7
2 1 10


Output

34


Input

7
1 2 3 5
1 1 10 10
1 4 5 -5
2 4 8
1 1 10 -10
2 4 8
2 1 10


Output

15
75
170

Note

In the first sample case, the danger value for x-coordinates 1 is 0 as there is no y2 satisfying the above condition for x = 1.

Danger values for x-coordinates 2 and 3 is 10 + | - 7| = 17.

Danger values for x-coordinates 4 to 9 is again 0 as there is no y2 satisfying the above condition for these coordinates.

Thus, total danger value is 17 + 17 = 34.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int INF = 1 << 30;
const int N = 100000;
int node[2][maxn << 2], se[2][maxn << 2], lazy[2][maxn << 2], cnt[2][maxn << 2];
long long sum[2][maxn << 2];
bool modify(int rt, int c, int k) {
  if (c <= se[k][rt]) return false;
  if (c < node[k][rt]) {
    sum[k][rt] -= (long long)(node[k][rt] - c) * cnt[k][rt];
    node[k][rt] = c;
    lazy[k][rt] = c;
  }
  return true;
}
void pushup(int rt) {
  for (int k = 0; k < 2; ++k) {
    node[k][rt] = max(node[k][rt << 1], node[k][rt << 1 | 1]);
    sum[k][rt] = sum[k][rt << 1] + sum[k][rt << 1 | 1];
    if (node[k][rt << 1] == node[k][rt << 1 | 1]) {
      se[k][rt] = max(se[k][rt << 1], se[k][rt << 1 | 1]);
      cnt[k][rt] = cnt[k][rt << 1] + cnt[k][rt << 1 | 1];
    } else if (node[k][rt << 1] < node[k][rt << 1 | 1]) {
      se[k][rt] = max(node[k][rt << 1], se[k][rt << 1 | 1]);
      cnt[k][rt] = cnt[k][rt << 1 | 1];
    } else {
      se[k][rt] = max(se[k][rt << 1], node[k][rt << 1 | 1]);
      cnt[k][rt] = cnt[k][rt << 1];
    }
  }
}
void pushdown(int rt) {
  for (int k = 0; k < 2; ++k) {
    if (lazy[k][rt] != INF) {
      modify(rt << 1, lazy[k][rt], k);
      modify(rt << 1 | 1, lazy[k][rt], k);
      lazy[k][rt] = INF;
    }
  }
}
void build(int rt, int l, int r) {
  for (int i = 0; i < 2; ++i) {
    sum[i][rt] = 0;
    cnt[i][rt] = 0;
    se[i][rt] = node[i][rt] = lazy[i][rt] = INF;
  }
  if (r - l == 1) return;
  int m = (l + r) >> 1;
  build(rt << 1, l, m);
  build(rt << 1 | 1, m, r);
}
void update(int rt, int l, int r, int x, int y, int c, int k) {
  if (c >= node[k][rt]) return;
  if (r - l == 1) {
    node[k][rt] = c;
    se[k][rt] = -INF;
    if (node[k ^ 1][rt] != INF) {
      sum[k][rt] = c;
      cnt[k][rt] = 1;
      sum[k ^ 1][rt] = node[k ^ 1][rt];
      cnt[k ^ 1][rt] = 1;
    }
    return;
  }
  if (x <= l && r <= y && modify(rt, c, k)) return;
  pushdown(rt);
  int m = (l + r) >> 1;
  if (x < m) update(rt << 1, l, m, x, y, c, k);
  if (m < y) update(rt << 1 | 1, m, r, x, y, c, k);
  pushup(rt);
}
long long query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return sum[0][rt] + sum[1][rt];
  pushdown(rt);
  int m = (l + r) >> 1;
  long long ret = 0;
  if (x < m) ret += query(rt << 1, l, m, x, y);
  if (m < y) ret += query(rt << 1 | 1, m, r, x, y);
  return ret;
}
int main() {
  build(1, 1, N);
  int q;
  scanf("%d", &q);
  while (q--) {
    int t, l, r, c;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d", &c);
      if (c > 0)
        update(1, 1, N, l, r, c, 0);
      else
        update(1, 1, N, l, r, -c, 1);
    } else
      printf("%I64d\n", query(1, 1, N, l, r));
  }
  return 0;
}
