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 int ST[410000];
int mark[410000];
long long int lazy[4100000];
int z = 0;
void construct(int pos, int begin, int end) {
  if (begin == end) {
    mark[pos] = begin;
    return;
  }
  int mid = (begin + end) / 2;
  construct(pos * 2, begin, mid);
  construct(pos * 2 + 1, mid + 1, end);
  mark[pos] = 0;
}
void add(int pos, int begin, int end) {
  if (mark[pos] > 0) {
    lazy[pos] += abs(mark[pos] - z);
    ST[pos] += 1LL * abs(mark[pos] - z) * (end - begin + 1);
  } else {
    int mid = (begin + end) / 2;
    add(pos * 2, begin, mid);
    add(pos * 2 + 1, mid + 1, end);
    ST[pos] =
        ST[pos * 2] + ST[pos * 2 + 1] + 1LL * lazy[pos] * (end - begin + 1);
  }
  mark[pos] = -1;
}
void update(int pos, int begin, int end, int l, int r) {
  if (begin > r || end < l) return;
  if (begin >= l && end <= r) {
    add(pos, begin, end);
    mark[pos] = z;
    return;
  }
  if (mark[pos] > 0) {
    mark[pos * 2] = mark[pos * 2 + 1] = mark[pos];
    mark[pos] = 0;
  }
  int mid = (begin + end) / 2;
  update(pos * 2, begin, mid, l, r);
  update(pos * 2 + 1, mid + 1, end, l, r);
  mark[pos] = 0;
  ST[pos] = ST[pos * 2] + ST[pos * 2 + 1] + 1LL * lazy[pos] * (end - begin + 1);
}
long long int find(int pos, int begin, int end, int l, int r) {
  if (begin >= l && end <= r) return ST[pos];
  int mid = (begin + end) / 2;
  if (r <= mid)
    return find(pos * 2, begin, mid, l, r) + 1LL * lazy[pos] * (r - l + 1);
  if (mid < l)
    return find(pos * 2 + 1, mid + 1, end, l, r) +
           1LL * lazy[pos] * (r - l + 1);
  return find(pos * 2, begin, mid, l, mid) +
         find(pos * 2 + 1, mid + 1, end, mid + 1, r) +
         1LL * lazy[pos] * (r - l + 1);
}
int main() {
  int i, n, m, type, x, y;
  scanf("%d %d", &n, &m);
  construct(1, 1, n);
  for (i = 1; i <= m; ++i) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d %d %d", &x, &y, &z);
      update(1, 1, n, x, y);
    }
    if (type == 2) {
      scanf("%d %d", &x, &y);
      printf("%I64d\n", find(1, 1, n, x, y));
    }
  }
}
