// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 100010;

int n;

ll m, ans = 1000000000000000000ll;

ll a[N];

struct bit {

	ll tree[N * 2];

	inline int lowbit(int i)

	{

		return i & -i;

	}

	void update(int pos, int num)

	{

		for(int i = pos; i <= 2 * m; i += lowbit(i)) 

			tree[i] += num;

	}

	ll query(int pos)

	{

		ll ret = 0;

		for(int i = pos; i; i -= lowbit(i)) 

			ret += tree[i];

		return ret;

	}

} t1, t2;

int main()

{

	scanf("%d%lld", &n, &m);

	for(int i = 1; i <= n; ++i) scanf("%lld", &a[i]);

	ll now = a[1], all = 0, to;

	for(int i = 2; i <= n; ++i)

	{

		to = a[i] - 1;

		now = a[i - 1] - 1;

		if(to < now) to += m;

		all += to - now;

		t1.update(now + 1, now);

		t1.update(to + 1, - now);

		t2.update(now + 1, 1);

		t2.update(to + 1, - 1);	

	}

	for(int i = 1; i <= m; ++i)

	{

		ll sum1 = t1.query(i), sum2 = t2.query(i), sum = all;

		sum -= (ll)i * sum2 - sum1 - sum2;

		sum1 = t1.query(i + m); 

		sum2 = t2.query(i + m);

		sum -= (ll)(i + m) * sum2 - sum1 - sum2;

		ans = min(ans, sum);  

	}

	printf("%lld\n", ans);

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<algorithm>

using namespace std;

#define ll long long

const int maxn=1e5+5;

const ll INF=0x3f3f3f3f3f3f3f3f;

int a[maxn];

struct node

{

    int l,r;

    ll diff,lz;

}tree[4*maxn];

void build(int l,int r,int cur)

{

    tree[cur].l=l;

    tree[cur].r=r;

    if(l==r)

        return;

    int mid=(l+r)>>1;

    build(l,mid,cur<<1);

    build(mid+1,r,cur<<1|1);

    return;

}

void push_up(int cur)

{

    tree[cur].diff=tree[cur<<1].diff+tree[cur<<1|1].diff;

    return;

}

void push_down(int cur)

{

    if(tree[cur].lz!=0)

    {

        tree[cur<<1].diff+=(tree[cur<<1].r-tree[cur<<1].l+1)*tree[cur].lz;

        tree[cur<<1].lz+=tree[cur].lz;

        tree[cur<<1|1].diff+=(tree[cur<<1|1].r-tree[cur<<1|1].l+1)*tree[cur].lz;

        tree[cur<<1|1].lz+=tree[cur].lz;

        tree[cur].lz=0;

    }

    return;

}

void add(int l,int r,ll val,int cur)

{

    if(l>r) return;

    if(tree[cur].l>=l&&tree[cur].r<=r)

    {

        tree[cur].diff+=(tree[cur].r-tree[cur].l+1)*val;

        tree[cur].lz+=val;

        return;

    }

    if(tree[cur].l>r||tree[cur].r<l)

        return;

    push_down(cur);

    add(l,r,val,cur<<1);

    add(l,r,val,cur<<1|1);

    push_up(cur);

    return;

}

ll diff0[maxn];

void together(int cur)

{

    if(tree[cur].l==tree[cur].r)

    {

        diff0[tree[cur].l]=tree[cur].diff;

        return;

    }

    push_down(cur);

    together(cur<<1);

    together(cur<<1|1);

    push_up(cur);

    return;

}

ll minn[maxn];

int main()

{

    int n,m;

    scanf("%d%d",&n,&m);

    for(int i=1;i<=n;i++)

        scanf("%d",&a[i]);

    build(0,m-1,1);

    ll base=0;

    for(int i=1;i<n;i++)

    {

        if(a[i]<a[i+1])

        {

            base+=(ll)(a[i+1]-a[i]);

            add(a[i]+1,a[i+1]-1,-1,1);

            add(a[i+1],a[i+1],a[i+1]-a[i]-1,1);

        }

        else

        {

            base+=(ll)(m-a[i]+a[i+1]);

            add(0,0,-(m-a[i]),1);

            add(1,a[i+1]-1,-1,1);

            add(a[i+1],a[i+1],m-a[i]+a[i+1]-1,1);

            add(a[i]+1,m-1,-1,1);

        }

    }

    together(1);

    minn[0]=base;

    for(int i=1;i<=m;i++)

        minn[i]=minn[i-1]+diff0[i-1];

    ll ans=INF;

    for(int i=1;i<=m;i++)

        ans=min(ans,minn[i]);

    printf("%lld\n",ans);

    return 0;

}

/*

 4 6

 1 5 1 4

 8 9 7 5 6 9

 

 9 9 8 7 6 5 4 3 2

 */


// Fastest known solution for this problem: