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 inf = 2000000000;
static inline int Rint() {
  struct X {
    int dig[256];
    X() {
      for (int i = '0'; i <= '9'; ++i) dig[i] = 1;
      dig['-'] = 1;
    }
  };
  static X fuck;
  int s = 1, v = 0, c;
  for (; !fuck.dig[c = getchar()];)
    ;
  if (c == '-')
    s = 0;
  else if (fuck.dig[c])
    v = c ^ 48;
  for (; fuck.dig[c = getchar()]; v = v * 10 + (c ^ 48))
    ;
  return s ? v : -v;
}
template <typename T>
static inline void cmax(T& a, const T& b) {
  if (b > a) a = b;
}
template <typename T>
static inline void cmin(T& a, const T& b) {
  if (b < a) a = b;
}
const int maxn = 100005;
long long delta[maxn << 2];
int mark[maxn << 2];
long long sum[maxn << 2];
void bc(int root, int l, int r) {
  if (l == r) {
    mark[root] = l;
    return;
  } else {
    const int lc = root << 1, rc = lc + 1, mid = l + r >> 1;
    bc(lc, l, mid);
    bc(rc, mid + 1, r);
  }
}
long long query(int root, int l, int r, int a, int b) {
  if (a == l && b == r) return sum[root];
  const int lc = root << 1, rc = lc + 1, mid = l + r >> 1;
  if (b <= mid) return query(lc, l, mid, a, b) + delta[root] * (b - a + 1);
  if (a > mid) return query(rc, mid + 1, r, a, b) + delta[root] * (b - a + 1);
  return query(lc, l, mid, a, mid) + query(rc, mid + 1, r, mid + 1, b) +
         delta[root] * (b - a + 1);
}
void push_down(int root, int l, int r, int x) {
  if (mark[root] > 0) {
    delta[root] += abs(mark[root] - x);
    sum[root] += 1LL * abs(mark[root] - x) * (r - l + 1);
  } else {
    const int lc = root << 1, rc = lc + 1, mid = l + r >> 1;
    push_down(lc, l, mid, x);
    push_down(rc, mid + 1, r, x);
    sum[root] = sum[lc] + sum[rc] + 1LL * delta[root] * (r - l + 1);
  }
  mark[root] = 0;
}
void update(int root, int l, int r, int a, int b, int x) {
  if (a == l && r == b) {
    push_down(root, l, r, x);
    mark[root] = x;
    return;
  }
  const int lc = root << 1, rc = lc + 1, mid = l + r >> 1;
  if (mark[root] > 0) {
    mark[lc] = mark[rc] = mark[root];
    mark[root] = 0;
  }
  if (b <= mid)
    update(lc, l, mid, a, b, x);
  else if (a > mid)
    update(rc, mid + 1, r, a, b, x);
  else
    update(lc, l, mid, a, mid, x), update(rc, mid + 1, r, mid + 1, b, x);
  sum[root] = sum[lc] + sum[rc] + 1LL * delta[root] * (r - l + 1);
}
int main() {
  const int n = Rint(), m = Rint();
  bc(1, 1, n);
  for (int i = (0); i < (m); ++i) {
    const int cmd = Rint(), l = Rint(), r = Rint();
    if (cmd == 1) {
      const int x = Rint();
      update(1, 1, n, l, r, x);
    } else {
      printf("%I64d\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}
