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 maxx = 1e5 + 100;
struct node {
  int l;
  int r;
  long long lazy;
  long long sum;
  long long cor;
} p[maxx << 2];
int n, m;
void pushup(int cur) {
  p[cur].cor = (p[cur << 1].cor == p[cur << 1 | 1].cor ? p[cur << 1].cor : 0);
  p[cur].sum = p[cur << 1].sum + p[cur << 1 | 1].sum;
}
void pushdown(int cur) {
  if (p[cur].lazy) {
    p[cur << 1].lazy += p[cur].lazy;
    p[cur << 1 | 1].lazy += p[cur].lazy;
    p[cur << 1 | 1].sum +=
        (p[cur].lazy * (p[cur << 1 | 1].r - p[cur << 1 | 1].l + 1));
    p[cur << 1].sum += (p[cur].lazy * (p[cur << 1].r - p[cur << 1].l + 1));
    p[cur << 1].cor = p[cur << 1 | 1].cor = p[cur].cor;
    p[cur].lazy = 0;
  }
}
void build(int l, int r, int cur) {
  p[cur].l = l;
  p[cur].r = r;
  p[cur].sum = p[cur].lazy = p[cur].cor = 0;
  if (l == r) {
    p[cur].cor = (long long)l;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, cur << 1);
  build(mid + 1, r, cur << 1 | 1);
  pushup(cur);
}
void update(int l, int r, int cur, long long add) {
  int L = p[cur].l;
  int R = p[cur].r;
  if (l <= L && R <= r && p[cur].cor) {
    p[cur].sum += (abs(add - p[cur].cor) * (R - L + 1));
    p[cur].lazy += (abs(add - p[cur].cor));
    p[cur].cor = add;
    return;
  }
  pushdown(cur);
  int mid = (L + R) / 2;
  if (r <= mid)
    update(l, r, 2 * cur, add);
  else if (l > mid)
    update(l, r, 2 * cur + 1, add);
  else {
    update(l, mid, 2 * cur, add);
    update(mid + 1, r, 2 * cur + 1, add);
  }
  pushup(cur);
}
long long query(int l, int r, int cur) {
  int L = p[cur].l;
  int R = p[cur].r;
  if (l <= L && R <= r) {
    return p[cur].sum;
  }
  pushdown(cur);
  int mid = (L + R) / 2;
  if (r <= mid)
    return query(l, r, 2 * cur);
  else if (l > mid)
    return query(l, r, 2 * cur + 1);
  else
    return query(l, mid, 2 * cur) + query(mid + 1, r, 2 * cur + 1);
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    build(1, n, 1);
    int x, y;
    long long z;
    int k;
    while (m--) {
      scanf("%d", &k);
      if (k == 1) {
        scanf("%d%d%lld", &x, &y, &z);
        update(x, y, 1, z);
      } else if (k == 2) {
        scanf("%d%d", &x, &y);
        printf("%lld\n", query(x, y, 1));
      }
    }
  }
  return 0;
}
