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 maxn = 2e5 + 10;
const long long inf = 1e9;
long long b[4 * maxn], lazy[4 * maxn], n, m, x, y, z, tree[maxn * 4];
char second[5];
void push_up(long long p) {
  b[p] = b[p * 2] + b[p * 2 + 1];
  tree[p] =
      ((tree[p * 2] && tree[p * 2 + 1]) && (tree[p * 2] == tree[p * 2 + 1]))
          ? tree[p * 2]
          : 0;
}
void push_down(long long p, long long l, long long r) {
  if (lazy[p] == 0) return;
  long long mid = (l + r) / 2;
  lazy[p * 2] += lazy[p];
  lazy[p * 2 + 1] += lazy[p];
  tree[p * 2] = tree[p];
  tree[p * 2 + 1] = tree[p];
  b[p * 2] += (mid - l + 1) * lazy[p];
  b[p * 2 + 1] += (r - mid) * lazy[p];
  lazy[p] = 0;
}
void build(long long p, long long l, long long r) {
  lazy[p] = 0;
  if (l == r) {
    tree[p] = l;
    b[p] = 0;
    return;
  }
  long long mid = (l + r) / 2;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  push_up(p);
}
void update(long long p, long long l, long long r, long long x, long long y,
            long long z) {
  if (x <= l && y >= r && tree[p]) {
    lazy[p] += abs(z - tree[p]);
    b[p] += (r - l + 1) * abs(z - tree[p]);
    tree[p] = z;
    return;
  }
  push_down(p, l, r);
  long long mid = (l + r) / 2;
  if (x <= mid) update(p * 2, l, mid, x, y, z);
  if (y > mid) update(p * 2 + 1, mid + 1, r, x, y, z);
  push_up(p);
}
long long Find(long long p, long long l, long long r, long long x,
               long long y) {
  long long ans = 0;
  if (x <= l && y >= r) return b[p];
  long long mid = (l + r) / 2;
  push_down(p, l, r);
  if (x <= mid) ans += Find(p * 2, l, mid, x, y);
  if (y > mid) ans += Find(p * 2 + 1, mid + 1, r, x, y);
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%lld", &x);
    if (x == 1) {
      scanf("%lld%lld", &x, &y);
      scanf("%lld", &z);
      update(1, 1, n, x, y, z);
    } else {
      scanf("%lld%lld", &x, &y);
      printf("%lld\n", Find(1, 1, n, x, y));
    }
  }
  return 0;
}
