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 N = 1e5 + 5;
struct SegTree {
  long long l, r, Min, Max;
  long long val, lazy;
} T[N << 2];
long long Abs(long long a) { return a < 0 ? -a : a; }
void pushup(int p) {
  T[p].Min = min(T[p << 1].Min, T[(p << 1) | 1].Min);
  T[p].Max = max(T[p << 1].Max, T[(p << 1) | 1].Max);
  T[p].val = T[p << 1].val + T[(p << 1) | 1].val;
}
void pushdown(int p) {
  if (T[p].l == T[p].r) {
    return;
  } else if (T[p].lazy) {
    T[p << 1].lazy += T[p].lazy;
    T[(p << 1) | 1].lazy += T[p].lazy;
    T[p << 1].val += T[p].lazy * (long long)(T[p << 1].r - T[p << 1].l + 1);
    T[(p << 1) | 1].val +=
        T[p].lazy * (long long)(T[(p << 1) | 1].r - T[(p << 1) | 1].l + 1);
    T[p << 1].Min = T[p << 1].Max = T[(p << 1) | 1].Max = T[(p << 1) | 1].Min =
        T[p].Min;
    T[p].lazy = 0;
  }
}
void build(int p, int l, int r) {
  int mid = (l + r) >> 1;
  T[p].l = l, T[p].r = r, T[p].lazy = 0;
  if (l == r) {
    T[p].Max = l, T[p].Min = l;
    T[p].val = 0;
    return;
  }
  build(p << 1, l, mid);
  build((p << 1) | 1, mid + 1, r);
  pushup(p);
}
void update(int p, int l, int r, long long add) {
  int mid = (T[p].l + T[p].r) >> 1;
  pushdown(p);
  if (l == T[p].l && T[p].r == r && T[p].Min == T[p].Max) {
    T[p].val += (long long)Abs(add - T[p].Min) * (long long)(r - l + 1);
    T[p].lazy += Abs(add - T[p].Max);
    T[p].Max = T[p].Min = add;
    return;
  }
  if (r <= mid) {
    update(p << 1, l, r, add);
  } else if (l > mid) {
    update((p << 1) | 1, l, r, add);
  } else {
    update(p << 1, l, mid, add);
    update((p << 1) | 1, mid + 1, r, add);
  }
  pushup(p);
}
long long query(int p, int l, int r) {
  int mid = (T[p].l + T[p].r) >> 1;
  pushdown(p);
  if (l == T[p].l && T[p].r == r) {
    return T[p].val;
  }
  if (r <= mid) {
    return query(p << 1, l, r);
  } else if (l > mid) {
    return query((p << 1) | 1, l, r);
  } else {
    return query(p << 1, l, mid) + query((p << 1) | 1, mid + 1, r);
  }
  pushup(p);
}
int main() {
  int n, m, l, r, c;
  long long add;
  scanf("%d %d", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%d %d %d", &c, &l, &r);
    if (c == 1) {
      scanf("%lld", &add);
      update(1, l, r, add);
    } else {
      printf("%lld\n", query(1, l, r));
    }
  }
  return 0;
}
