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;
long long n, m;
const int N = 1e5 + 5;
struct node {
  long long l, r;
  long long lz;
  long long sum;
  long long maxx;
  long long minn;
} tree[N << 2];
long long absll(long long a) {
  if (a < 0)
    return -a;
  else
    return a;
}
void pushup(int i) {
  tree[i].minn = min(tree[i << 1].minn, tree[i << 1 | 1].minn);
  tree[i].maxx = max(tree[i << 1].maxx, tree[i << 1 | 1].maxx);
  tree[i].sum = tree[i << 1].sum + tree[i << 1 | 1].sum;
}
void pushdown(int i) {
  if (tree[i].l == tree[i].r)
    return;
  else if (tree[i].lz) {
    tree[i << 1].lz += tree[i].lz;
    tree[i << 1 | 1].lz += tree[i].lz;
    long long mid = (tree[i].l + tree[i].r) >> 1;
    tree[i << 1].sum += tree[i].lz * (mid - tree[i << 1].l + 1);
    tree[i << 1 | 1].sum += tree[i].lz * (tree[i << 1 | 1].r - mid);
    tree[i << 1].minn = tree[i << 1].maxx = tree[i << 1 | 1].minn =
        tree[i << 1 | 1].maxx = tree[i].minn;
    tree[i].lz = 0;
  }
}
void build(int i, int l, int r) {
  int mid = (l + r) >> 1;
  tree[i].l = l;
  tree[i].r = r;
  tree[i].lz = 0;
  if (l == r) {
    tree[i].maxx = tree[i].minn = l;
    tree[i].sum = 0;
    return;
  }
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  pushup(i);
}
void updata(int i, int l, int r, long long k) {
  int mid = (tree[i].l + tree[i].r) >> 1;
  pushdown(i);
  if (l == tree[i].l && r == tree[i].r && tree[i].minn == tree[i].maxx) {
    tree[i].sum += absll(k - tree[i].minn) * (long long)(r - l + 1);
    tree[i].lz += absll(k - tree[i].minn);
    tree[i].maxx = tree[i].minn = k;
    return;
  }
  if (r <= mid) {
    updata(i << 1, l, r, k);
  } else if (l > mid) {
    updata(i << 1 | 1, l, r, k);
  } else {
    updata(i << 1, l, mid, k);
    updata(i << 1 | 1, mid + 1, r, k);
  }
  pushup(i);
}
long long search(int i, int l, int r) {
  if (tree[i].l >= l && tree[i].r <= r) return tree[i].sum;
  if (tree[i].r < l || tree[i].l > r) return 0;
  pushdown(i);
  long long s = 0;
  if (tree[i * 2].r >= l) s += search(i * 2, l, r);
  if (tree[i * 2 + 1].l <= r) s += search(i * 2 + 1, l, r);
  return s;
}
int main() {
  scanf("%lld %lld", &n, &m);
  build(1, 1, n);
  while (m--) {
    int a, l, r;
    scanf("%d %d %d", &a, &l, &r);
    if (a == 1) {
      long long k;
      scanf("%lld", &k);
      updata(1, l, r, k);
    } else {
      printf("%lld\n", search(1, l, r));
    }
  }
  return 0;
}
