DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int AX = 1e5 + 66;
int tot;
struct Node {
  int col;
  long long sum;
  long long lazy;
} s[AX << 2];
void pushUp(int rt) {
  s[rt].sum = s[rt << 1].sum + s[rt << 1 | 1].sum;
  if (s[rt << 1].col == s[rt << 1 | 1].col) {
    s[rt].col = s[rt << 1].col;
  } else
    s[rt].col = 0;
  return;
}
void pushDown(int rt, int L, int R) {
  if (s[rt].col && s[rt].lazy) {
    int mid = (L + R) >> 1;
    s[rt << 1].col = s[rt].col;
    s[rt << 1 | 1].col = s[rt].col;
    s[rt << 1].sum += (mid - L + 1) * s[rt].lazy;
    s[rt << 1 | 1].sum += (R - mid) * s[rt].lazy;
    s[rt << 1].lazy += s[rt].lazy;
    s[rt << 1 | 1].lazy += s[rt].lazy;
    s[rt].lazy = 0;
  }
  return;
}
void build(int l, int r, int rt) {
  if (l == r) {
    s[rt].col = ++tot;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushUp(rt);
}
void update(int L, int R, int v, int l, int r, int rt) {
  if (L <= l && R >= r && s[rt].col) {
    s[rt].sum +=
        (long long)((long long)(r - l + 1) * (long long)abs(v - s[rt].col));
    s[rt].lazy += abs(v - s[rt].col);
    s[rt].col = v;
    return;
  }
  pushDown(rt, l, r);
  int mid = (l + r) >> 1;
  if (L <= mid) update(L, R, v, l, mid, rt << 1);
  if (R > mid) update(L, R, v, mid + 1, r, rt << 1 | 1);
  pushUp(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return s[rt].sum;
  }
  pushDown(rt, l, r);
  long long ans = 0LL;
  int mid = (l + r) >> 1;
  if (L <= mid) ans += query(L, R, l, mid, rt << 1);
  if (R > mid) ans += query(L, R, mid + 1, r, rt << 1 | 1);
  pushUp(rt);
  return ans;
}
int main() {
  int n, m;
  int op;
  int l, r, x;
  scanf("%d%d", &n, &m);
  tot = 0;
  build(1, n, 1);
  while (m--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &l, &r, &x);
      update(l, r, x, 1, n, 1);
    } else {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", query(l, r, 1, n, 1));
    }
  }
  return 0;
}
